describe gravitation using a field model and apply Newton's law of universal gravitation $F = G m_1 m_2 / r^2$ and the relationships $g = G M / r^2$, $g = F/m$, the work done by a gravitational field $W = \Delta U = mg \Delta h$ in a uniform field and the change in gravitational potential energy in non-uniform fields as the area under a force-distance graph

What this dot point is asking

VCAA wants you to describe gravity using a field model, apply Newton's law of universal gravitation, calculate field strength $g$, and find changes in gravitational potential energy both in a uniform field ($mg\Delta h$) and in a non-uniform field (the area under the $F$ vs $r$ graph).

The answer

The field model

A gravitational field surrounds every mass. A second mass placed in the field experiences a force toward the first. The field is a vector at each point, drawn with field lines that point in the direction of the force on a positive test mass.

• Near a planet's surface the field is approximately uniform (parallel field lines, $g \approx 9.8$ N/kg downward).
• At large distances the field is radial (lines pointing inward toward the planet) and the strength falls as $1/r^2$.

Newton's law of universal gravitation

Every pair of masses attracts with a force:

where $G = 6.67 \times 10^{-11}$ N m squared per kg squared and $r$ is the centre-to-centre distance.

Field strength

The gravitational field strength $g$ at a point is the force per unit mass:

It is also numerically equal to the acceleration of a freely falling object (because $F = mg$ and $F = ma$ together give $a = g$). Units are N/kg or m/s squared.

At Earth's surface, $g = 9.8$ N/kg. The field strength falls with altitude as $1/r^2$ where $r$ is measured from Earth's centre.

Gravitational potential energy in a uniform field

Close to a planet's surface, the field is approximately uniform. The change in gravitational potential energy when an object of mass $m$ is raised through height $\Delta h$ is:

This is the work done against gravity.

Gravitational potential energy in a non-uniform field

Far from a planet's surface, the field varies with distance and the equation $mg\Delta h$ no longer applies. The change in potential energy between two distances $r_1$ and $r_2$ equals the work done against the gravitational force, which is the area under the $F$ vs $r$ graph between those points.

VCAA expects you to use the graphical area for non-uniform fields. The analytic equivalent is:

for moving from $r_1$ to $r_2 > r_1$ (positive when moving outward, because work is done against gravity).

Field lines and energy: a quick visual rule

• Field lines point in the direction of the force on a positive test mass (always toward the planet).
• A mass moved with the field (downward) loses potential energy.
• A mass moved against the field (outward) gains potential energy.

Worked example with numbers

A satellite of mass $500$ kg is at altitude $300$ km above Earth's surface. Find the gravitational force on it and the field strength at that point. ($M_E = 6.0 \times 10^{24}$ kg, $R_E = 6.4 \times 10^6$ m.)

$r = 6.4 \times 10^6 + 3.0 \times 10^5 = 6.7 \times 10^6$ m.

$g = \frac{G M}{r^2} = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{(6.7 \times 10^6)^2} = \frac{4.00 \times 10^{14}}{4.49 \times 10^{13}} = 8.9$ N/kg.

$F = m g = 500 \times 8.9 = 4.45 \times 10^3$ N.

Try it: Universal gravitation calculator - enter the two masses and separation, or use the Kepler third-law calculator for satellite orbits.

Common traps

Using $mg\Delta h$ for changes in altitude of hundreds of kilometres. That equation assumes a uniform field; over a 300 km altitude change, $g$ drops by about 9 percent. Use the area under the $F$ vs $r$ graph instead.

Forgetting that $r$ is measured from the centre of the planet. An altitude of 600 km above a planet of radius 6400 km gives $r = 7000$ km.

Confusing field strength with force. Field strength $g$ has units N/kg and is independent of the test mass. The force $F = mg$ depends on the test mass.

Treating the sign of $\Delta U$ casually. When an object moves outward (away from the planet), $\Delta U$ is positive (the system gains potential energy). When it falls inward, $\Delta U$ is negative.

Forgetting that gravitational force is always attractive. Field lines always point inward toward the source mass.

In one sentence

A gravitational field is a vector field with strength $g = GM/r^2$ surrounding every mass, exerting a force $F = G m_1 m_2 / r^2$ on other masses; changes in potential energy are $mg\Delta h$ in a uniform field and the area under the $F$ vs $r$ graph in a non-uniform field.